Question

$selected_subcats =  EstablishmnetSubcategories::Join('establishments','establishments.id','=','establishment_subcategories.establishment_id')->where('establishment_subcategories.created_by',Auth::User()->id)->get(array('establishment_subcategories.status as sub_status'));

这是我的控制器代码及其输出

[{"status":1},{"status":1},{"status":1},{"status":1},{"status":1},{"status":1},{"status":1},{"status":0}]

我有一个条件，例如，我需要从控制器检查状态，我有一个条件，如至少一个状态值应为0（active），我想要前端的jquery验证应提醒我至少一个状态应单击按钮时为0（有效）。谁能帮忙，如何完成..谢谢

下面的代码是我的视图代码

<td><a href="javascript:void;"  class="label label-warning">{{$p->sub_status == 0 ? '<span class="label label-warning">Unpublish </span>' : '<span class="label label-warning">publish </span>'}}</a></td>

Answer 1

您需要修改查询以返回状态为0（有效）的类别数

$activeCats =  EstablishmnetSubcategories::Join('establishments','establishments.id','=','establishment_subcategories.establishment_id')
->where('establishment_subcategories.created_by',Auth::User()->id)
->where('establishment_subcategories.status',0)
->count();

html：

<td><a href="javascript:void;"  class="label label-warning">{{$activeCats  > 0 ? '<span class="label label-warning">Unpublish </span>' : '<span class="label label-warning">publish </span>'}}</a></td>

从控制器获取值并显示警报，至少应激活一个值

1 个答案: